Download CHAPtER 2 Energy calculations

2
CH A P T E R
Energy calculations
FS
You will examine:
meaning of enthalpy and the ΔH notation
profile diagrams for exothermic and endothermic
reactions
■■ thermochemical equations and subsequent calculations
that can be made using them
■■ the calculation of the ΔH value for a reaction and from
experimental data
■■ the kinetic molecular theory of gases
■■ the concept of gas pressure and common units used to
measure it
■■ the absolute (or Kelvin) temperature scale
■■ Boyle’s law and Charles’ law
■■ standard conditions for measuring gas volumes and
molar volume
■■ the universal gas equation
■■ mass–volume and volume–volume stoichiometry
■■ the calculation of greenhouse gas amounts associated
with the production of energy from carbon-based fuels
■■ how ΔH values may be estimated from simple
laboratory apparatus.
■■ the
PR
O
O
■■ energy
U
N
C
O
R
R
EC
TE
D
PA
G
E
‘Energy is neither created nor destroyed; it can only
be converted from one form to another’; this is one of
the famous laws of thermodynamics. Energy powers
the way we live, but where does this energy come
from? The most concentrated form of energy is held
in the bonds of chemicals we live with every day.
How can we determine the best choices for fuels to
produce this energy? How can we make quantitative
predictions that match energy requirements to fuel
quantities?
Many modern homes and buildings use ‘burning log’ fireplaces
for their heating. Such fireplaces do not burn logs at all — they
merely give the illusion that they are. In many cases, their heat
output comes from fuels such as ethanol, natural gas and LPG.
In designing such fireplaces, it is important to know the energy
output of the chosen fuel so that combustion can produce the
required amount of energy. It is also important to control the fuel
to oxygen ratio so that unwanted and dangerous combustion
products, such as carbon monoxide, are avoided.
Energy changes in chemical reactions
TE
D
A change in enthalpy occurs
when magnesium burns.
PA
G
E
PR
O
O
FS
The energy changes that accompany chemical reactions are vital to us. To
survive, we depend on the energy content of the food we eat. Our bodies can
convert the energy of the chemical bonds in food into other kinds of energy.
The quality of lifestyle we lead depends on harnessing energy from different
chemical sources, including coal, oil, natural gas and renewable fuels. The
study of the energy changes that accompany chemical reactions is called
thermochemistry or chemical thermodynamics. In general, all chemical
reactions involve energy changes. The chemical energy stored in a substance
has the potential to be converted to heat (which will be discussed in this
chapter) or electricity (see chapter 3).
A certain amount of chemical energy is stored within every atom, molecule
or ion. This energy is the sum of the potential energy and kinetic energy of the
substance and results from:
• the attractions and repulsions present between protons and electrons within
the atom
• the attractions and, to some degree, repulsions present between atoms
within a molecule
• the motion of the electrons
• the movement of the atoms.
The total energy stored in a substance is called the enthalpy, or heat content,
of the substance and is given the symbol H. Unfortunately we cannot directly
measure the heat content of a substance. What we can measure is the
change in enthalpy when the substance undergoes a chemical reaction. Since,
in virtually all chemical reactions, the energy of the reactants and products
differ, such reactions usually involve some change in enthalpy, which is indicated by a temperature rise or fall. The change in enthalpy during a reaction is
known as the heat of reaction and is denoted by ∆H.
Let us consider the following reaction:
2Mg(s) + O2(g)
2MgO(s)
O
R
R
EC
Magnesium has a certain enthalpy. So too does the oxygen. The total enthalpy
of the reactants — magnesium and oxygen — can be represented by the
symbol Hr. The product, magnesium oxide, also has some enthalpy — call
this Hp.
We can say that:
U
N
C
Enthalpy is the heat content of a
substance. It cannot be measured
directly, but the change in
enthalpy, known as the heat of
reaction, ∆H, can be measured.
Enthalpy changes are stated at
SLC; they refer to the energy
needed to return the temperature
to 25 °C and the pressure to
100.0 kPa.
32
Unit 3
change in heat content = (heat content of products) – (heat content of reactants)
So, enthalpy change is equal to heat energy produced or absorbed.
∆H = Hp – Hr
The enthalpy change, or heat change, of a reaction depends on the amount
of reactants used and the temperature of the initial reactants, compared with
that of the final products. To remove these variables in enthalpy studies, the
following conditions are assumed:
• the initial temperature of the reactants and final temperature of the products
is the same, and is 25 °C
• 1 mole of the substance is involved
• solutions have a concentration of 1 M
• the heat absorbed or released is measured in kilojoules (kJ)
• the pressure is kept at 100.0 kPa.
These conditions are referred to as standard laboratory conditions (SLC).
Although the usual name for ∆H is heat of reaction (or change in enthalpy),
there are some reactions for which specific names have been given.
• Heat of solution is the change in enthalpy when 1 mole of any substance dissolves in water.
• Heat of neutralisation is the change in enthalpy when an acid reacts with a
base to form 1 mole of water.
• Heat of vaporisation is the change in enthalpy when 1 mole of liquid is converted to a gas.
• Heat of combustion is the enthalpy change when a substance burns in air,
and is always exothermic.
FS
Exothermic and endothermic reactions
O
Chemical reactions accompanied by heat energy changes may be divided into
two groups: exothermic reactions and endothermic reactions.
E
PR
O
Exothermic reactions
Chemical reactions that release heat to the environment are called exothermic
reactions.
MgCl2(aq) + H2(g)
heat energy
given out;
feels warm
magnesium
chloride
solution
D
Mg(s) + 2HCl(aq)
PA
hydrochloric acid
G
magnesium
EC
When a strip of magnesium is placed in hydrochloric acid, heat is given out.
This means that the enthalpy of the reactants, Hr , is greater than the enthalpy
of the products, Hp. Therefore, the enthalpy change, ∆H = Hp – Hr , is negative.
The difference in energy between the reactants and the products is released
into the environment, usually in the form of heat.
The energy change in a reaction can be drawn as an energy diagram or
energy profile. Chemicals with more energy are drawn higher up and those
with less energy lower down. The reactants are drawn on the left and the
products are drawn on the right. The figure below is an energy diagram for an
exothermic reaction.
N
C
O
R
R
Exothermic reactions release
heat. The enthalpy change, ∆H,
is negative.
TE
An exothermic reaction — magnesium ribbon dissolving in hydrochloric acid
reactants
energy
U
Endothermic reactions absorb
heat. The enthalpy change, ∆H,
is positive.
ΔH = energy change = heat of reaction
products
In an exothermic reaction, ∆H is always negative because the heat content of
the reactants is greater than that of the products. The bonds in the products are
more stable than the bonds in the reactants.
CHAPTER 2 Energy calculations
33
Endothermic reactions
Chemical reactions that absorb heat from the environment are called
endothermic reactions.
ammonium
nitrate
water
An endothermic reaction —
ammonium nitrate dissolving
in water
FS
heat energy
taken in; feels
cold
ammonium
nitrate
solution
PA
energy
Digital document
Experiment 2.1 Investigating heat
changes in reactions
doc-18833
G
E
PR
O
O
When ammonium nitrate is dissolved in water, heat is absorbed from the
environment; the test tube feels cold because the reaction takes in energy from
the water and your hand. This means that the enthalpy of the products, Hp, is
greater than the enthalpy of the reactants, Hr. Therefore, the enthalpy change,
∆H = Hp – Hr, is positive. The difference in energy between the reactants and
the products is absorbed from the environment. The figure below is an energy
diagram for an endothermic reaction.
products
ΔH = energy change = heat of reaction
D
reactants
EC
TE
In an endothermic reaction, ∆H is always positive because the heat content of
the reactants is less than that of the products. The bonds in the products are less
stable than the bonds in the reactants.
Revision question
U
N
C
O
R
R
1. State whether the following are exothermic or endothermic processes.
(a) Water changing from liquid to gaseous state
(b)A reaction in which the total enthalpy of the products is greater than
that of the reactants
(c) Burning kerosene in a blow torch
(d) Burning kerosene in a jet aircraft engine
(e) A chemical reaction that has a negative ΔH value
(f) A reaction in which the reactants are at a lower level on an energy
profile diagram than the products
A thermochemical equation is a
balanced equation that includes
the amount of heat produced or
absorbed.
34
Unit 3
Thermochemical equations
As we saw briefly in chapter 1, an equation that includes the amount of heat
produced or absorbed by a reaction is a thermochemical equation. As with
other chemical equations, charge and mass must balance. However, they must
also include the enthalpy change.
In writing a thermochemical equation, the following points should be
remembered:
• A positive or negative sign must be included with the ∆H value to indicate
whether the reaction is either endothermic or exothermic. If an enthalpy
change is given as ∆H = 345 kJ mol–1, the lack of sign does not mean that it
is an endothermic reaction.
• Since enthalpy is measured in the units kJ mol–1, the coefficients in the
equation represent the amount in moles of each reacting substance. So, the
reaction
2H2(g) + O2(g)
2H2O(l) ∆H = –572 kJ mol–1
Concept 3
Do more
Combustion
equations
In one case, the product is a liquid. In the other case, the product is a gas. So,
the condensation of 2 moles of water vapour to 2 moles of liquid water at 25 °C
produces 88 kJ of energy, which is the difference between the two enthalpies.
• If the coefficients are doubled, the ∆H value must be doubled.
O
Topic 1
2H2(g) + O2(g)
4H2(g) + 2O2(g)
PR
AOS 1
2H2O(l) ∆H = –572 kJ mol–1
2H2O(g) ∆H = –484 kJ mol–1
O
2H2(g) + O2(g)
2H2(g) + O2(g)
Unit 3
FS
can be read as: when 2 moles of hydrogen react with 1 mole of oxygen,
2 moles of water form and 572 kJ of energy is produced. ∆H refers to the
equation as it is written, even though the unit is expressed as kJ mol–1.
• The physical state of matter must be shown, since changes of state require
energy changes.
2H2O(l) ∆H = –572 kJ mol–1
4H2O(l) ∆H = –1144 kJ mol–1
2H2O(l)
PA
G
E
The amount of energy produced by a chemical reaction is directly proportional
to the amount of substance initially present. If, as in the above example, twice
as much reactant is used, then twice as much energy can be produced.
• If a reaction is reversed, ∆H is equal to, but opposite in sign, to that of the
forward reaction.
2H2(g) + O2(g) ∆H = +572 kJ mol–1
EC
TE
D
You will notice that the key point is that the enthalpy change in a reaction is
proportional to the amount of substance that reacts. If these two quantities are
measured in an experiment, it is possible to write the accompanying thermochemical equation. This was covered in chapter 1. Sample problem 2.1 provides another example of how this can be done.
Sample problem 2.1
O
R
R
In an experiment, it was found that the combustion of 0.240 g of methanol
in excess oxygen yielded 5.42 kJ. Write the thermochemical equation for this
reaction.
The equation for this reaction is:
2CO2(g) + 4H2O(l)
2CH3OH(l) + 3O2(g)
U
N
C
Solution:
To match this equation, it is necessary to calculate the heat evolved from
2 moles (64.0 g) of methanol. This can be done by scaling up the results from
the experiment as follows.
64.0
× 5.42 = 1.45 × 103 kJ
Heat evolved from 2 moles (64.0 g) =
0.240
The thermochemical equation is therefore:
2CH3OH(l) + 3O2(g)
2CO2(g) + 4H2O(l) ΔH = –1.45 × 103 kJ mol–1
Sample problem 2.2
The molar heat of combustion of ethanol is tabulated as –1364 kJ mol–1.
(a) Write the thermochemical equation for the combustion of ethanol.
(b) If the density of ethanol is 0.790 g mL–1, calculate the energy evolved in MJ
when 1.00 L of ethanol is burned.
CHAPTER 2 Energy calculations
35
(a) The balanced chemical equation for the combustion of ethanol is:
2CO2(g) + 3H2O(l)
CH3CH2OH(l) + 3O2(g)
The given heat value refers to 1 mole, as does the equation above.
The thermochemical equation is therefore:
2CO2(g) + 3H2O(l) ΔH = –1364 kJ mol–1
CH3CH2OH(l) + 3O2(g)
mass
(b) Since density =
volume
mass of ethanol burned (m) = d(ethanol) × V(ethanol)
= 0.790 × 1000
= 790 g
790
= 17.17 mol
n(ethanol) =
46.0
From the equation, 1 mole of ethanol evolves 1364 kJ.
By direct proportion, 17.17 moles produce x kJ.
1364
Therefore, x = 17.17 ×
= 23 425 kJ = 23.4 MJ.
1
PR
O
O
FS
Solution:
E
Sample problem 2.3
G
The combustion of ethene can be represented by the following thermochemical
equation.
2CO2(g) + 2H2O(l) ΔH = –1409 kJ mol–1
PA
C2H4(g) + 3O2(g)
Calculate the mass of ethene required to produce 500 kJ of heat energy.
From the equation, 1 mole of C2H4 evolves 1409 kJ.
By ratio, x moles are required to evolve 500 kJ.
1
= 0.355 mol
x = 500 ×
1409
Therefore, m(C2H4) = 0.355 × 28.0 = 9.94 g.
EC
TE
D
Solution:
Sample problem 2.4
U
N
C
O
R
R
The air pollutant sulfur trioxide reacts with water in the atmosphere to produce
sulfuric acid according to the equation:
SO3(g) + H2O(l)
H2SO4(aq) ∆H = –129.6 kJ mol–1
Calculate the energy released, in kJ, when 0.500 kg SO3(g) reacts with water.
n(SO3) =
Solution:
m 500
=
= 6.24 mol
M 80.1
From the equation, 1 mole of SO3 releases 129.6 kJ.
So, 6.24 moles release 6.24 × 129.6 = 809 kJ of heat energy.
Revision questions
2. Calculate the energy released, in kJ, when 3.56 g of carbon undergoes combustion according to the thermochemical equation:
2C(s) + O2(g)
2CO(g) ∆H = –788 kJ mol–1
3. The use of hydrogen as a renewable and environmentally friendly fuel is
currently the subject of much research. The main product of hydrogen
36
Unit 3
combustion is water. The production of liquid water from the reaction
between gaseous hydrogen and gaseous oxygen can be represented by the
following thermochemical equation.
O2(g) + 2H2(g)
2H2O(l) ∆H = –572 kJ mol–1
Calculate how much energy, in kilojoules, would be released or absorbed by
the following reactions.
(a)2O2(g) + 4H2(g)
4H2O(l)
(b)H2O(l)
1
O (g)
2 2
+ H2(g)
FS
4. During the production of sulfuric acid by the Contact process, sulfur dioxide
is converted to sulfur trioxide according to the equation:
2SO2(g) + O2(g)
2SO3(g) ∆H = –197 kJ mol–1
Calculate the heat energy released in the production of 1.00 tonne (106 g) of
sulfur trioxide gas.
5. Calculate the energy released when 18.5 g of carbon undergoes combustion
in a plentiful supply of air according to the equation:
C(s) + O2(g)
CO2(g) ∆H = –394 kJ mol–1
PR
O
O
6. Methanol burns according to the equation:
2CH3OH(l) + 3O2(g)
2CO2(g) + 4H2O(l) ΔH = –1450 kJ mol–1
PA
G
E
Calculate the mass of methanol required to produce 1.000 MJ of energy.
7. Butane and octane are two hydrocarbons commonly used as fuels. The
thermochemical equations for these two fuels are shown below.
2C4H10(g) + 13O2(g)
8CO2(g) + 10H2O(l) ΔH = –5748 kJ mol–1
2C8H18(g) + 25O2(g)
16CO2(g) + 18H2O(l) ΔH = –10 928 kJ mol–1
TE
D
(a) Calculate the heat evolved by the combustion of 100 g of butane.
(b) Use your answer to (a) to calculate the mass of octane required to produce the same amount of energy.
8. Explain why it is critical to show symbols of state in thermochemical
equations.
If the enthalpy change of a desired reaction is not known, it is possible to calculate it from a series of related reactions. The calculated value may then be
used to make thermochemical predictions as shown previously. To do this, use
some of the dot points mentioned on pages 34–35. The following sample problems show how this is done.
N
C
O
R
R
The ΔH value of a reaction can
be calculated by manipulating
and then adding together two or
more related reactions and their
associated ΔH values.
EC
Calculating ΔH values from two or more
related reactions
Sample problem 2.5
U
Calculate the enthalpy change for the incomplete combustion of methane in a
limited oxygen supply, given the following two equations.
CH4(g) + 2O2(g)
2CO(g) + O2(g)
Solution:
CO2(g) + 2H2O(l) ΔH = –889 kJ mol–1(equation 1)
2CO2(g) ΔH = –556 kJ mol–1
(equation 2)
The required equation is 2CH4(g) + 3O2(g)
2CO(g) + 4H2O(l) (equation 3)
Equations 1 and 2 need to be ‘manipulated’ so that, when added together,
equation 3 is produced. This can be achieved by multiplying equation 1 by 2
and reversing equation 2.
2CH4(g) + 4O2(g)
2CO2(g)
2CO2(g) + 4H2O(l) ΔH = –1778 kJ mol–1
2CO(g) + O2(g) ΔH = +556 kJ mol–1
CHAPTER 2 Energy calculations
37
Adding together results in cancelling to produce the required equation. The
ΔH values are also added, thus producing a value of –1778 + 556 = –1222.
2CO(g) + 4H2O(l) ΔH = –1222 kJ mol–1.
Therefore, 2CH4(g) + 3O2(g)
Sample problem 2.6
Fossil fuels such as coal and petroleum contain sulfur as an impurity. This produces sulfur oxides when they are burned. These are atmospheric pollutants.
Develop the thermochemical equation for the reaction:
S(s) + O2(g)
FS
SO2(g)
2SO3(g) ΔH = –196 kJ mol–1
2SO2(g) + O2(g)
2SO3(g) ΔH = –760 kJ
mol–1
(equation 1)
(equation 2)
PR
O
2S(s) + 3O2(g)
Equation 1 needs to be halved and reversed. Equation 2 needs to be halved.
The ΔH values thus become +98 and –380 respectively. Adding the adjusted
equations together produces the required thermochemical equation.
SO2(g) ΔH = –282 kJ mol–1
E
S(s) + O2(g)
PA
Revision questions
G
Solution:
O
given the following equations.
 9. Calculate the ΔH value for the incomplete combustion of pentane, according
to the equation:
10CO(g) + 12H2O(l)
given the following equations.
TE
D
2C5H12(l) + 11O2(g)
2CO2(g) ΔH = –556 kJ mol–1
2CO(g) + O2(g)
C5H12(l) + 8O2(g)
5CO2(g) + 6H2O(l) ΔH = –3509 kJ mol–1
given the following equations.
CH3CH2OH(l) + 3O2(g)
U
N
C
O
R
The kinetic molecular theory
states that matter is made up of
continuously moving particles.
38
Unit 3
CH3CH2OH(l) + 3O2(g)
R
EC
10.Calculate ΔH for the reaction:
H2O(l)
2CO2(g) + 3H2O(g)
2CO2(g) + 3H2O(l) ΔH = –1364 kJ mol–1
H2O(g) ΔH = +44.0 kJ mol–1
Working with gases
When considering fuels as energy sources, it is often necessary to deal with
gases. The products of combustion are nearly always gases, and some of the
fuels themselves are gases. While it is possible to measure their masses, it is
often more convenient and meaningful to measure their volumes.
Gases have properties and exhibit behaviours that are different from liquids
and solids. Scientists use the kinetic molecular theory of gases to explain
observed gas properties. This consists of five postulates (or points) that form a
mental picture of how the particles in a gas would look and behave if we could
observe them directly. These are applicable to gas samples under ‘moderate
conditions’, which is usually taken to mean pressures that are not much greater
than atmospheric pressure and temperatures considerably greater than those
at which the gases liquefy.
PR
O
O
FS
Hot air balloons use the
properties of gases to achieve
flight. A large fan is used
to blow air into the balloon.
Once inflated, the air is heated
using a propane burner. The
molecules move more rapidly,
hitting each other in a random
chaotic motion. As they move
further apart, the density of
the air decreases. The balloon
rises because the air inside is
warmer, and therefore lighter,
than the surrounding air.
TE
D
PA
G
E
The five postulates that make up the kinetic molecular theory of gases can
be summarised as follows.
1. Gases are made up of particles moving constantly and at random.
2. Gas particles are very far apart, and the volume of the particles is very small
compared with the volume that the gas occupies.
3. The forces of attraction and repulsion between gas particles are practically
zero.
4.Gas particles collide with each other and the walls of their container,
exerting pressure. The collisions are perfectly elastic. This means that no
kinetic energy is lost when they collide.
5. The higher the temperature, the faster the gas particles move, as they have
increased kinetic energy.
EC
Specifying the large-scale behaviour
of gases
O
R
R
In any consideration of gas behaviour, the pressure each gas exerts, the volume
that it occupies, its temperature and the number of gas molecules present in the
sample must be determined.
Gas pressure
Particles exert a force by colliding with the walls of a container. Each tiny collision
adds to all the others to make up the continuous force that we call pressure.
U
N
C
Pressure is the force exerted by
gas particles on the walls of a
container.
The surface of an inflatable
airbed, such as the one shown
here, exerts a force and tries
to collapse. For the airbed
to stay inflated, the particles
inside the airbed must be
able to exert a large enough
force to balance the forces
exerted by the surface and the
external air pressure.
CHAPTER 2 Energy calculations
39
G
E
PR
O
O
FS
Pressure is force per unit area.
Pressure (P) is defined as the force exerted per unit area.
force
P=
area
Pressure is measured using various units.
The SI unit of pressure is the pascal (Pa), where 1 pascal is equivalent to a
force of 1 newton exerted over an area of 1 square metre (N m–2).
EC
Atmospheric pressure
The pressure of the atmosphere is measured by a barometer. Atmospheric
pressure at sea level is 101 325 Pa. This is usually simplified to read 101.3 kPa
(kilopascal). Two older units of pressure are millimetres of mercury and atmospheres. One millimetre of mercury (1 mmHg) is defined as the pressure needed
to support a column of mercury 1 mm high. This unit developed from the early
use of mercury barometers. One atmosphere (1 atm) is the pressure required to
support 760 mm of mercury (760 mmHg) in a mercury barometer at 25 °C. This
is the average atmospheric pressure at sea level. So, 1 atm equals 760 mmHg.
760 mmHg = 1.00 atm = 1.013 × 105 Pa = 101.3 kPa
In weather reports, bars and hectopascals are used to measure gas pressure.
1 bar = 100 000 Pa = 100 kilopascal (kPa)
1
bar = 100 Pa = 1 hectopascal (hPa)
1 millibar (mb) =
1000
U
N
C
O
R
R
Units of pressure: 760 mmHg =
1.00 atm = 101.3 kPa
TE
D
PA
If a tennis racquet is struck by
a tennis ball, a small force is
felt. If a stream of balls is fired
at the racquet, it is felt as a
continuous pressure.
Units of temperature:
0 °C = 273 K
40
Unit 3
Temperature
The Celsius scale takes the freezing point of water as 0 °C and the boiling point
of water as 100 °C. The space between these two fixed points is divided into
100 equal intervals, or degrees. Temperatures below the freezing point of water
are assigned negative values, such as –10 °C.
Another temperature scale is the Kelvin, or absolute, scale. On the Kelvin
scale, the freezing point of water is 273 K and its boiling point is 373 K. Notice
that a change of 1° on the Celsius scale is the same as that on the Kelvin
scale. The zero point on the Kelvin scale, 0 K or absolute zero, is –273 °C. The
100 °C
373 K
relationship between the temperature on the Celsius scale and that on the
Kelvin scale is given by the following equations.
K = °C + 273 or °C = K – 273
For example, to convert 25 °C to the absolute scale:
100
divisions
K = 25 + 273
= 298
100
divisions
Note that temperatures given in K do not have a ° sign.
FS
Volume
Revision question
11. Make the following conversions.
(a) 780 mmHg to atm
(e) 200 °C to K
(b) 4.0 atm to Pa
(f) 500 K to °C
(c) 1000 mmHg to Pa
(g) 3.0 m3 to L
(d) 1250 mmHg to kPa
(h) 250 L to mL
PA
(i) 1600 mL to L
(j) 3 × 106 mL to L
(k) 5 × 103 mL to m3
(l) 600 mL to m3
D
Units of volume:
1 m3 = 103 L = 106 mL
O
1 m3 = 103 L = 106 mL
G
These thermometers show a
comparison of the Celsius and
Kelvin temperature scales.
or
1 m3 = 1000 L = 1 000 000 mL
PR
273 K
E
0 °C
O
The volume of a gas is commonly measured in cubic metres (m3) or litres (L)
or millilitres (mL).
TE
Laws to describe the behaviour of gases
The behaviour of gases has been studied for centuries. As a result, a number of
laws have evolved to describe their behaviour mathematically. These laws are
independent of the type of gas — it does not matter what the gas is (or even if
the gas is a mixture such as air), these laws apply in exactly the same way to all
gases. Two very useful such laws are Boyle’s law and Charles’ law.
U
N
C
O
R
R
EC
Digital document
Experiment 2.2 The relationship
between pressure and volume
doc-18834
Boyle’s law for pressure–volume changes
Boyle’s law is named after the English physicist and chemist who discovered
the relationship between pressure and volume for a sample of gas. It states
that, for a fixed amount of gas at constant temperature, pressure is inversely
proportional to volume. Mathematically, this can be represented as:
P∝
1
V
from which it can be stated that PV = a constant value.
If P1V1 = a constant value and P2V2 = a constant value, then:
P1V1 = P2V2
Revision questions
12. (a)Set up a spreadsheet and enter the data from the table on the next page
into the cells of the first two columns.
(b) In the third column of the spreadsheet, calculate pressure × volume.
What pattern do you notice?
CHAPTER 2 Energy calculations
41
(c) In the fourth column of your spreadsheet, convert each pressure
 1
. For example, if P has a value
 P 
measurement (P) into its reciprocal 
1
would equal 8.33 × 10–3.
 P 
13. (a)Use the data on your spreadsheet from question 12 to plot a graph of
pressure versus volume. Put pressure on the horizontal (x) axis and
volume on the vertical (y) axis.
1
(b) Plot a second graph with the volume on the vertical axis and on the
P
horizontal axis, and extrapolate the graph to the origin.
(c) Compare and account for the two graphs.
FS
of 120, 
Sample results showing the relationship between the pressure and volume of a gas
Volume (L)
120
0.261
O
O
Pressure (kPa)
145
0.218
PR
162
180
200
0.120
PA
258
0.159
0.130
G
240
0.171
0.145
E
216
0.193
–50
0.75
0
0.92
40
1.05
70
1.15
120
1.32
180
1.52
U
Volume (L)
2
Volume (L)
N
C
Temperature (°C)
O
R
R
Table 2.1 Sample results
showing the relationship
between the temperature and
volume of a gas
EC
TE
D
Charles’ law for temperature–volume changes
The relationship between temperature and volume was first identified by the
French scientist Jacques Charles, after whom the law is named. While the
expansion of all materials with increasing temperature is well known, Charles
subjected constant amounts of various gases (at constant pressure) to changes
in temperature, each time making accurate measurements of the resulting
volume.
To understand this law better, consider a typical set of results as shown in
table 2.1.
If these results are graphed as shown below, a pattern emerges.
1
0
−273
100
200
300
400
500
K
−173
−73
27
127
227
°C
Temperature (°C and K)
Representation of Charles’ law showing the two temperature scales, Kelvin and
degrees Celsius. The graph demonstrates that the volume of a gas varies in
direct proportion to absolute temperature.
42
Unit 3
FS
If the Celsius scale is used to measure temperature, a linear relationship is
observed. However, this is not a directly proportional relationship as the graph
does not pass through the origin. If, however, the Kelvin scale is used, it does.
We can therefore state Charles’ law as follows.
For a given amount of gas at constant pressure, volume is directly proportional to the absolute temperature.
Mathematically, this can be represented as, V ∝ T , (where T is the absolute
V
temperature), from which it can be stated that = a constant value.
T
V
V
If 1 = a constant value and 2 = a constant value, then:
T1
T2
O
V1 V2
=
T1 T2
PR
O
You will notice that the graph on the previous page implies a ‘lowest possible temperature’ — the temperature at which it intersects the x-axis. This temperature is –273.15 °C or 0 K. It is referred to as ‘absolute zero’.
The combined gas equation
E
Boyle’s law and Charles’ law may be combined to produce the gas equation:
G
P1V1 P2V2
=
T1
T2
D
PA
This equation applies to a fixed amount of gas. It is particularly useful for making
predictions when an amount of gas under a set of initial conditions (P1, V1 and T1)
is changed to a second set of conditions (P2, V2 and T2). If any two of the new set
of conditions are known, the third can easily be calculated.
TE
Sample problem 2.7
EC
A sample of air at 10 °C and 1.1 atm has a volume of 2.2 L. If the pressure is
changed to 1.0 atm and the temperature increases to 15 °C, what volume will
it occupy?
Solution:
Step 1
U
N
C
O
R
R
List the known information:
P1 = 1.1 atm
V1 = 2.2 L
T1 = 10 + 273 = 283 K
P2 = 1.0 atm
V2 = ? L
T2 = 15 + 273 = 288 K
Step 2
Transpose the general gas equation and substitute the values above.
P1V1 P2V2
=
T1
T2
PV T
V2 = 1 1 2
P2T1
1.1 × 2.2 × 288
=
1.0 × 283
= 2.5 L
The new volume is, therefore, 2.5 L.
Note that units for pressure and volume must be the same on each side.
Temperature must be in kelvin.
CHAPTER 2 Energy calculations
43
Standard conditions for
measuring gases
FS
PR
O
O
STP and SLC are standard sets of
conditions for comparing gases.
Boyle’s law and Charles’ law tell us that the volume of a gas sample is sensitive
to both temperature and pressure. This makes the comparison of gas volumes
very tricky. In order to make these comparisons easier, scientists have established sets of standard conditions. These are accepted worldwide and enable
gas volumes to be compared meaningfully without temperature or pressure
having unwanted effects.
There are two commonly used such sets of standard conditions.
1. Standard temperature and pressure (STP). Standard temperature is 0 °C
or 273 K, and standard pressure is 100.0 kPa.
2. Standard laboratory conditions (SLC). Since most experiments are
carried out in the laboratory, normal room temperature is substituted for
the STP condition. Standard laboratory conditions are a temperature of
25 °C (298 K) and a pressure of 100.0 kPa.
This book will generally refer to SLC only, as STP is not part of the VCE
Chemistry study design.
G
E
Molar gas volume and Avogadro’s
hypothesis
PA
Avogadro put forward the hypothesis that ‘equal volumes of all gases
measured at the same temperature and pressure contain the same number
of particles’.
This means that, if two gases have the same temperature, pressure and
volume, they must contain the same number of moles.
It has been found experimentally that 1 mole of any gas at SLC occupies a
volume of 24.8 L. This volume is called the molar gas volume.
EC
TE
D
The molar gas volume is the
volume of 1 mole of any gas at a
stated temperature and pressure.
VM (STP) = 22.7 L
VM (SLC) = 24.8 L
oxygen
U
N
C
O
R
R
nitrogen
V = 24.8 L
T = 298 K
P = 100.0 kPa
6.02 × 1023 molecules
V = 24.8 L
T = 298 K
P = 100.0 kPa
6.02 × 1023 molecules
The standard molar gas volume. Under the same conditions of temperature
and pressure, the volume of a gas depends only on the number of molecules
it contains, and not on what the particles are. The volume occupied by
28.0 g of nitrogen gas (1 mol or 6.02 × 1023 molecules) at SLC is 24.8 L.
The volume occupied by 1 mole of any gas at the same temperature and
pressure is 24.8 L. (Note: In these diagrams, the number of particles shown is
only a fraction of those actually present, since this number is too great to be
represented here.)
44
Unit 3
The molar volume of a gas at SLC is 24.8 L. This means that 1 mole of any gas
occupies 24.8 L at 25 °C and 100.0 kPa. The molar volume of a gas varies with
temperature and pressure but, at any given temperature and pressure, it is the
same for all gases. There is a direct relationship between the number of moles
of a gas and its molar volume. This is illustrated by the formula
n=
V
Vm
O
FS
where n is the number of moles of gas, V is the actual volume and Vm is the
molar volume of the gas. Therefore, at SLC:
V
nSLC =
24.8
where V is measured in L.
O
Sample problem 2.8
V
24.8
n=
48.0
= 1.94 mol of Cl 2 gas
24.8
E
n=
PA
Sample problem 2.9
G
Solution:
PR
How many moles of chlorine gas are present in 48.0 L of the gas at SLC?
Find the mass of 1556.5 mL of H2 gas that was collected at SLC.
O
R
R
Why are these bubbles
floating upwards? These
soap bubbles are filled with
hydrogen gas, which is
lighter than the same volume
of bubbles filled with air. A
mole of hydrogen gas would
occupy the same volume as a
mole of oxygen molecules, but
the hydrogen would not weigh
as much as the oxygen.
EC
TE
D
Solution:
Step 1
Convert the volume to litres.
1556.5 mL = 1.5565 L
Step 2
Determine the number of moles of H2 gas
V
using n =
24.8
1.5565
24.8
= 0.0628 mol
n=
Step 3
U
N
C
Determine the mass (m) of 0.0628 moles of
m
H2 gas using n(H2 ) =
M
m
0.0628 =
2.0
The mass of H2 gas collected was 0.13 g.
Sample problem 2.10
(a) 5345 mL of a gas was collected at SLC and weighed. Its mass was 9.50 g.
Find the molar mass of the gas.
(b) Given that the gas is one of the main constituents of air, identify the gas.
Solution:
(a) STEP 1
Convert the volume to litres.
5345 mL = 5.345 L
CHAPTER 2 Energy calculations
45
Step 2
Determine the number of moles of the gas.
V
n=
24.8
5.345
=
24.8
= 0.216 mol
Step 3
Determine the molar mass (M) of 0.216 mole of the gas.
m
n=
M
9.50
0.216 =
M
9.50
M=
0.216
= 44.0
The molar mass of the gas is 44.0 g mol–1.
(b) The main constituents of air are nitrogen (M = 28.0), oxygen (M = 32.0),
argon (M = 40.0) and carbon dioxide. The gas must therefore be carbon
dioxide as the M of CO2 = 12 + (2 × 16) = 44.0 g mol–1.
PA
Revision questions
G
E
PR
O
O
FS
N
C
O
R
R
EC
TE
D
14. Calculate the numbers of moles of the following gases at SLC.
(a) 15 L of oxygen, O2
(b) 25 L of chlorine, Cl2
15. Calculate the volumes of the following gases at SLC.
(a) 1.3 mol hydrogen, H2
(b) 3.6 g of methane, CH4
(c) 0.35 g of argon, Ar
16. Calculate the masses of the following gas samples. All volumes are measured at SLC.
(a) 16.5 L of neon, Ne
(b) 1050 mL of sulfur dioxide, SO2
17. What is the mass (in kg) of 850 L of carbon monoxide gas measured at SLC?
18. A 0.953 L quantity of a gas measured at SLC has a mass of 3.20 g. What is the
molar mass of the gas? What is the gas?
The universal gas equation
U
PV = nRT
is used to calculate the amount
of a gas at a given volume and
temperature.
It is important that the correct
units are used in the universal gas
equation.
46
Unit 3
The universal gas equation
When Boyle’s law and Charles’ law are combined with Avogadro’s hypothesis,
the following equation is obtained.
PV = nRT
This equation is known as the general gas equation or the universal gas
equation.
When using the universal gas equation, the following units must be used for
the universal gas constant R = 8.31 J K–1 mol–1.
• Pressure is measured in kilopascals.
• Volume is measured in litres.
• Temperatures is measured in kelvin.
• The quantity of gas is measured in moles
• The number of moles can be expressed as:
m
n=
M
where m represents the mass of the substance (in grams) and M represents the
molar mass of the substance.
So, the universal gas equation can be rewritten as:
PV =
mRT
M
FS
Sample problem 2.11
Topic 1
Concept 4
O
PA
Step 2
PR
Do more
The mole
E
Assign the variables.
P = 550 kPa
V=?L
n = 6.30 mol
T = 23 + 273 = 296 K
R = 8.31 K–1 mol–1
Unit 3
AOS 1
Step 1
G
Solution:
O
Find the volume of 6.30 mol of carbon dioxide gas at 23.0 °C and 550 kPa pressure. (R = 8.31 J K–1 mol–1)
Use the universal gas equation to calculate V.
EC
TE
D
PV = nRT
nRT
∴V =
P
6.30 × 8.31 × 296
=
550
= 28.2 L
Revision questions
U
N
C
O
R
R
19. Calculate the volume of gas, in litres, occupied by the following.
(a) 3.5 mol of O2 at 100 kPa and 50 °C
(b) 12.8 mol of CH4 at 10 atm and 60 °C
(c) 6.5 g of Ar at 50 kPa and 100 °C
(d) 0.56 g of CO2 at 50 atm and 20 °C
(e) 1.3 × 10–3 g He at 60 kPa and –50 °C
(f) 1.5 × 1021 molecules of Ne at 40 kPa and 200 °C
20. Calculate the volume occupied by 42.0 g of nitrogen gas at a pressure of
200 kPa and a temperature of 77 °C.
21. A 5.00 L balloon contains 0.200 mol of air at 120 kPa pressure. What is the
temperature of air in the balloon?
22. If 55 mol of H2 gas is placed in a 10 L flask at 7 °C, what would be the pressure
in the flask?
Mass – volume stoichiometry
Quantities of gases can be
calculated using the four
stoichiometric steps.
Many chemical reactions involve gases. For example, barbecues, furnaces
and engines, such as the internal combustion engine, burn fuel and produce
carbon dioxide and water vapour. In order to ensure that the reactions are
CHAPTER 2 Energy calculations
47
efficient and as complete as possible, calculations of volumes or masses of the
gases or fuels required are essential.
Sample problem 2.12
In a gas barbecue, propane is burned in oxygen to form carbon dioxide and
water vapour. If 22.0 g of CO2 is collected and weighed, find the volume of propane at 200 °C and 1.013 × 105 Pa.
PR
Step 2
O
Write the equation and given information.
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
5
P = 1.013 × 10 Pa m(CO2) = 22.0 g
T = 200 + 273 = 473 K
V=?L
FS
Step 1
O
Solution:
22.0
= 0.500 mol
44.0
G
n(CO2 ) =
E
Calculate the number of moles of the known quantity of substance.
Step 3
PA
Use the equation to find the molar ratio of unknown to known quantities, and
calculate the number of moles of the required substance. Since 3 moles of CO2
are produced from 1 mole of C3H8
n(CO2 )
3
0.500
=
3
= 0.167 mol
EC
TE
D
n(C 3H8 ) =
Step 4
U
N
C
O
R
R
Find the volume of the propane using PV = nRT, where R = 8.31 J K–1 mol–1
and P = 101 325 Pa = 101.325 kPa. Then,
V(C3H8) = nRT = 0.167 × 8.31 × 473
P
101.3
V(C3H8) = 6.47 L
Revision questions
23. Ethyne, C2H2, is also called acetylene and is used as a welding gas. It is produced from calcium carbide as shown by the ethyne reaction.
CaC2(s) + 2H2O(l)
Ca(OH)2(s) + C2H2(g)
What volume of ethyne measured at 25 °C and 745 mmHg would be produced from 5.00 g of H2O?
24. The equation for photosynthesis to produce glucose is as follows.
48
Unit 3
6CO2(g) + 6H2O(g)
C6H12O6(g) + 6O2(g)
What volume of carbon dioxide, measured at 28 °C and 1.3 atm pressure, is
needed to produce 1.5 g of glucose?
Sample problem 2.13
Calculate the mass of hexane that is required to produce 100 L of carbon
dioxide gas at SLC.
Solution:
Step 1
Write the equation and identify the given information.
2C6H14(l) + 19O2(g)
12CO2(g) + 14H2O(l)
m = ? g V = 100 L Conditions are SLC.
FS
Step 2
PR
O
O
Calculate the number of moles of the known substance — in this case, CO2 gas.
Normally the universal gas equation would be used but, because it is at SLC,
we know the molar volume and can therefore use the formula
V
n=
Vm
100
= 4.03 mol.
Therefore, n(CO2) =
24.8
Step 3
PA
G
E
Using the mole ratios from the equation, the number of moles of hexane can
be calculated.
2
n(C6H14) =
× n(CO2)
12
2
× 4.03 = 0.672 mol.
Therefore, n(C6H14) =
12
Step 4
D
The mass of hexane can now be calculated.
TE
m(C6H14) = n(C6H14) × M(C6H14) = 0.672 × 86.0 = 57.8 g
EC
Revision questions
25. Oxygen gas can be prepared in the laboratory by the decomposition of potassium nitrate according to the equation:
U
N
C
O
R
R
2KNO3(s)
2KNO2(s) + O2(g)
When 14.5 L of O2 is formed at 1.00 atm and 25.0 °C, what mass of KNO2 is
also formed?
26. Magnesium reacts with hydrochloric acid according to the equation:
Mg(s) + 2HCl(aq)
MgCl2(aq) + H2(g)
What mass of magnesium, when reacted with excess hydrochloric acid, would
produce 5.0 L of hydrogen gas, measured at 26.0 °C and 1.2 kPa pressure?
If gases are at the same pressure
and temperature, their molar ratios
are equal to their volume ratios.
N2(g) + 3H2(g)
2NH3(g)
1 mol 3 mol 2 mol
1 vol
3 vol 10 vol 30 vol 2 vol
20 vol
Volume – volume stoichiometry
Consider the reaction between nitrogen and hydrogen gas.
N2(g) + 3H2(g)
2NH3(g)
According to this equation, 1 mole of N2 reacts with 3 moles of H2 to produce
2 moles of ammonia.
Considering Avogadro’s hypothesis, if the gases are at the same pressure and
temperature, their molar ratios are equal to their volume ratios. We therefore
use volumes instead of moles and can say that 10 mL of N2 reacts with 30 mL of
H2 to form 20 mL of ammonia.
CHAPTER 2 Energy calculations
49
Sample problem 2.14
If 100 m3 of ethene is burned according to
2CO2(g) + 2H2O(g)
C2H4(g) + 3O2(g)
calculate the volume of:
(a) carbon dioxide produced
(b) oxygen consumed.
(Assume all gas volumes are measured at the same temperature and pressure.)
Because all gas volumes are measured at the same temperature and pressure,
the equation may be interpreted in terms of volume ratios.
(a) V(CO2) = 2 × V(C2H4)
Therefore, V(CO2) = 2 × 100 = 200 m3.
(b) V(O2) = 3 × V(C2H4)
Therefore, V(O2) = 3 × 100 = 300 m3.
PR
O
O
FS
Solution:
Revision questions
E
27. Methane gas burns in air at room temperature and pressure, according to
the equation:
D
PA
G
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
If 25 mL of methane is burned at room temperature and pressure, find the
volumes of the following reactants and products:
(a) oxygen
(b) carbon dioxide
(c) water.
28. At high temperatures, such as those in a car engine during operation, atmospheric nitrogen burns to produce the pollutant nitrogen dioxide, according
to the equation:
TE
N2(g) + 2O2(g)
2NO2(g)
U
N
C
O
R
R
EC
(a) If 20 mL of nitrogen is oxidised, calculate the volume of oxygen needed
to produce the pollutant. Assume that temperature and pressure remain
constant.
(b) What is the initial volume of reactants in this combustion reaction?
(c) What is the final volume of products in the reaction?
(d) Is there an overall increase or decrease in the volume of gases on completion of the reaction?
Applying volume stoichiometry to
thermochemistry
Just as the principles of mass–mass stoichiometry can be extended to include
gas volumes, the thermochemical calculations introduced earlier in this
chapter can also be extended to include gas volumes. This is very useful
because it is often more convenient to measure a gas’s volume than its mass.
If pressure and temperature are measured as well, the universal gas equation,
pV = nRT, provides a formula to change between volumes and moles.
Sample problem 2.15
Calculate the heat energy released when 375 mL of methane at 21 °C and
767 mmHg pressure is burned according to:
CH4(g) + 2O2(g)
50
Unit 3
CO2(g) + 2H2O(l) ΔH = –889 kJ mol–1
The universal gas equation,
PV = nRT, can be used in
thermochemical calculations to
link energy amounts to volumes of
gases involved.
Step 1
Calculate the number of moles of methane used. To do this, the information
in the question needs to be changed to the appropriate units to enable the
universal gas equation to be used.
Temperature: 21 °C = 21 + 273 = 294 K
Volume: 375 mL = 0.375 L
767
Pressure: 767 mmHg =
× 101.3 = 102.2 kPa
760
PV 102.2 × 0.375
=
= 0.0157 mol.
Therefore, n(CH4) =
8.31 × 294
RT
FS
Solution:
Step 2
O
O
0.0157
= 14.0 kJ
1
PR
Heat evolved = 889 ×
Revision questions
C6H12O6(aq) + 6O2(g) ΔH = +2816 kJ mol–1
G
6CO2(g) + 6H2O(l)
E
29. Calculate the energy required to convert 2.50 L of carbon dioxide, at SLC, to
glucose according to the equation:
PA
30. Calculate the energy released when 200 mL of diborane, B2H6, is burned at
150 °C and 1.50 atmospheres according to the following equation.
B2O3(s) + 3H2O(g) ΔH = –2035 kJ mol–1
D
B2H6(g) + 3O2(g)
TE
Fuels and greenhouse gases
U
N
C
O
R
R
EC
The principles of stoichiometry and thermochemistry can be combined to
quantify the effect that energy production, especially through the combustion
of fossil fuels, has on the environment. As we saw in chapter 1, gases such as
carbon dioxide, methane, nitrous oxide and ozone contribute to the enhanced
greenhouse effect. Of these, carbon dioxide is produced in the greatest
amounts due to the use of fossil fuels as energy sources. The increasing level of
these gases has led to international treaties such as the Kyoto Protocol. These
obligate signatory countries to reduce their greenhouse gas emissions in an
attempt to limit future global warming.
In order to meet targets for reduction, countries need to estimate their energy
requirements, both current and future, and compare the energy sources available for the required energy production. A useful unit in such comparisons is
the amount of greenhouse gas produced (usually carbon dioxide) per megajoule of energy released. Typical units are g MJ–1 and L MJ–1.
Sample problem 2.16
If it is assumed that petrol is entirely octane, and that it burns according to
2C8H18(l) + 25O2(g)
16CO2(g) + 18H2O(l) ΔH = –10 928 kJ mol–1
calculate the:
(a) mass of carbon dioxide produced per MJ of energy evolved
(b) volume of carbon dioxide produced at 100 kPa and 20 °C per MJ of energy
evolved.
CHAPTER 2 Energy calculations
51
(a) Step 1
Calculate the mass of octane required to evolve 1 MJ of heat energy. From
the equation, 2 moles of octane evolves 10.928 MJ.
M(C8H18) = 114.0 g mol–1
Therefore, 2 × 114.0 g evolves 10.928 MJ.
2 × 114.0
Therefore, 1 MJ requires
= 20.9 g.
10.928
Step 2
Calculate the mass of CO2 produced from this amount of octane.
FS
Solution:
m 20.9
=
= 0.183 mol
M 114.0
16
From the equation, n(CO2) =
× 0.183 = 1.464 mol.
2
Therefore, m(CO2) = 1.464 × 44.0 = 64.4 g.
It can therefore be stated that, if petrol is assumed to be pure octane, it
g of carbon dioxide per megajoule of energy released
produces 64.4 (64.4 g MJ–1).
(b) PV = nRT
n = 1.464 mol
T = 273 + 20 = 293 K
P = 100 kPa
nRT 1.464 × 8.31 × 293
Therefore, V =
=
= 35.6 L.
P
100
Therefore, 35.6 L of CO2 is produced per megajoule of energy released
(35.6 L MJ–1) at the stated conditions of temperature and pressure.
G
E
PR
O
O
n(C8H18) =
TE
D
PA
When comparing the effect of
fuels on the environment, it is
often useful to consider the carbon
dioxide emissions produced per
unit of energy. Two common units
for making such comparisons are
g MJ–1 and L MJ–1.
EC
Sample problem 2.17
U
N
C
O
R
R
Methane is produced by a number of natural processes and, in many of these
situations, it can easily enter the atmosphere. Methane is a more potent greenhouse gas than carbon dioxide as it is better at absorbing heat. However, it can
also be used as a fuel. The equation for burning methane is:
Solution:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
If methane is captured and then used as a fuel, calculate the net change
in greenhouse gas volume that would occur. Assume that all volumes are
measured at the same temperature and pressure.
As temperature and pressure is constant, we can write:
V(CH4) consumed = V(CO2) produced
Therefore, there is no net change.
You might think that a process such as capturing and then burning methane
would produce no net environmental benefit. However, when it is remembered
that methane is a more potent greenhouse gas than carbon dioxide, it can be
seen that there is an environmental benefit gained from burning the methane,
rather than just letting it escape into the atmosphere. Of course, if carbon
dioxide could also be captured and prevented from entering the atmosphere,
the environmental benefit would be even greater.
Carbon dioxide is the most significant greenhouse gas due to the large volumes
in which it is produced. For this reason, other greenhouse gas emissions are often
52
Unit 3
measured in ‘carbon dioxide equivalent’. That is, all these other gases are converted to the volume of carbon dioxide that would produce the same effect.
Revision question
FS
31. (a)Calculate the net change in mass of greenhouse gas produced by the
combustion of 128 g of methane.
(b) Express your answer as a percentage increase or decrease.
(c) Comment on how the units used (mass or volume) may influence the
conclusions drawn from calculations such as in this question and in
sample problem 2.17.
U
PA
D
TE
EC
N
C
O
R
R
Both transport and the
generation of electricity
consume large amounts
of fossil fuels and, in the
process, add large amounts
of carbon dioxide to the
atmosphere. As nations work
to reduce their greenhouse
gas emissions, alternative
fuels need to be considered.
When considering fuels,
the amount of carbon
dioxide emitted per unit of
energy produced is a useful
comparison.
G
E
PR
O
O
When electricity generation is considered, the calculations must take into
account the efficiency of the generation process. For example, to produce 1 MJ
of electrical energy, enough fuel would need to be burned to produce 3.33 MJ
of heat energy, assuming that the process is 30% efficient. The mass of carbon
dioxide produced in supplying 3.33 MJ would then need to be calculated.
Solution:
Sample problem 2.18
Calculate the carbon dioxide emissions, in kg MJ–1, from a power station that
has an overall efficiency of 33.0% and burns black coal with a carbon content
of 81.5%. (Heat evolved from black coal = 34.0 kJ g–1)
At 33.0% efficiency, the energy input required to produce 1 MJ of electrical
energy output is:
output energy
× 100
Efficiency =
input energy
1 × 100
= 3.03 MJ.
33
3030
Mass of black coal required to produce 3.03 MJ =
= 89.1 g
34.0
81.5
× 89.1 = 72.6 g
Mass of carbon in this amount of black coal =
100
Therefore, input energy =
CHAPTER 2 Energy calculations
53
Assuming the carbon produces carbon dioxide according to:
C(s) + O2(g)
CO2(g)
72.6
n(C) =
= 6.05 mol and n(C) = n(CO2)
12.0
Therefore, m(CO2) = 6.05 × 44.0 = 267 g = 0.267 kg.
Therefore, CO2 production is 0.267 kg MJ–1 of electrical energy produced.
Revision questions
PA
G
E
PR
O
O
FS
32. The molar heat of combustion for ethanol is –1364 kJ mol–1. Calculate the mass
of carbon dioxide emitted when ethanol is used to produce 1.00 MJ of heat.
33. (a)Methanol is also a fuel. Its molar heat of combustion is –725 kJ mol–1.
What mass of carbon dioxide would be produced using methanol to
generate 1.00 MJ of heat?
(b) Comment on your answers to questions 32 and 33(a) in relation to the
masses of CO2 produced.
34.A coal-fired power station using brown coal as its fuel operates at 37.0%
overall energy efficiency. The brown coal has a heat value of 16.0 kJ g–1 and a
carbon content of 29.0%. Assuming that all the carbon present forms carbon
dioxide, calculate the carbon dioxide produced per MJ of electrical energy
produced in units of:
(a) g MJ–1
(b) L MJ–1 (at SLC).
D
How do we obtain the energy output of a fuel?
EC
TE
So far in this chapter, we have made predictions relating to fuel quantities and
heat outputs using thermochemical equations. However, there are two issues
that you may have been wondering about.
• How are ΔH values obtained in the first place?
• Thermochemical equations are fine for fuels that are pure substances and
can therefore have an equation written for their combustion. But what about
fuels that are mixtures? Many common fuels, such as petrol, diesel, wood
and coal, fall into this category. How can we obtain heat outputs and make
predictions for these fuels?
The answer to both these questions is to burn a small amount of fuel and
‘capture’ the heat evolved in some way so that it can be measured. The results
obtained are then scaled up to a reference amount so that comparisons can be
made. The values obtained are quoted in units that take into account the units
used to measure the energy output and the particular reference amount being
used. Common units are kJ mol–1, kJ g–1 and MJ tonne–1.
If this is to be done accurately, a device called a calorimeter must be used.
There are two main types of calorimeters — solution calorimeters and bomb
calorimeters. The features of these and their use are discussed in chapter 12.
An alternative method is to use the fuel to heat a known mass of water
and measure the resultant temperature increase. This method produces only
approximate results, the degree of accuracy depending on the steps taken to
minimise heat loss to the surroundings.
You may remember from unit 1 that the formula:
N
C
O
R
R
The energy available from a fuel
may be estimated by using a
known amount of fuel to heat
a known mass of water. If the
temperature change is noted, the
formula Q = mcΔT may be used to
calculate the energy added to the
water. Knowing the amount of fuel
used, this can then be scaled up to
an appropriate reference amount.
Common units are kJ g–1, kJ mol–1
and MJ tonne–1.
Unit 3
U
AOS 1
Topic 1
Concept 7
Do more
Heats of
combustion
Q = mcΔT
can be used to calculate the heat required to raise a given mass (m) of a substance of known specific heat (c) by a certain temperature (ΔT). The specific
54
Unit 3
heat capacity, c, of a substance is the amount of energy needed to raise the
temperature of 1 g of the substance by 1 °C or 1 K. This formula is used to
calculate the heat added to the water. Sample problem 2.19 illustrates this
method.
Sample problem 2.19
Step 1
O
Solution:
FS
The heat content of kerosene was determined by using a kerosene burner to
heat 250 mL of water. It was found that burning 0.323 g of kerosene raised the
temperature of the water by 11.2 °C.
Given that the specific heat capacity of water is 4.18 J g–1 K–1, calculate the
heat energy released from kerosene in kJ g–1 and MJ tonne–1.
PR
O
Calculate the heat energy (Q) transferred to the water.
250 mL of water = 250 g
Q = mcΔT = 250 g × 4.18 J g K–1 × 11.2 °C
= 11 704 J = 11.7 kJ
Step 2
TE
D
PA
G
E
Scale this value up to the chosen reference amount (in this case, 1 g).
1
Heat evolved = 11.7 kJ ×
= 36.2 kJ g–1
0.323 g
To convert this to MJ tonne–1, 1 tonne = 106 g = 1 000 000 g; 1 MJ = 1000 kJ
36.2
Therefore, 36.2 kJ g–1 =
× 1 000 000 = 36 200 MJ tonne–1.
1000
Note: This represents a minimum value. Not all of the heat from the kerosene
would have gone into the water. Some of it would have been wasted in heating
the surrounding air and the equipment used to hold the water.
EC
Revision questions
U
N
C
O
R
R
An experiment to compare the energy output of candle wax, ethanol and butane
was performed by setting up the apparatus shown in the figure at the top of the
next page. The ethanol was poured into a crucible, a small wax candle stuck onto
a watch glass and a gas cigarette lighter were each used as a ‘burner’ after being
lit. Each burner was weighed before and after it was used to heat 200 g of water.
The results are shown in table 2.2 on the next page. (Assume the formula for
candle wax is C20H42.)
35. Copy and complete table 2.2.
36. Taking the specific heat capacity of water as 4.18 J g–1 °C−1, calculate the energy
produced from 1.00 gram of each substance. How do the results compare?
37. Calculate the heat of combustion (enthalpy per mole of substance used) for
each substance. How do the results compare?
38. Describe a use for each of the fuels tested. Explain why each fuel is better suited
for the purpose it is commonly used for than either of the other two fuels.
39. The molar heat of combustion of ethanol has been found experimentally to
be –1364 kJ mol−1.
(a) Is combustion exothermic or endothermic?
(b) What was the percentage accuracy of the experiment above?
(c) List the sources of error in the experiment and describe how some of
these errors can be minimised.
40. Write a report for the experiment.
CHAPTER 2 Energy calculations
55
thermometer
retort stand
clamp
copper can
heatproof mats
used to keep
out draughts
water
FS
fuel
(candle wax)
O
fuel container
(watch glass)
O
safety mat
E
Table 2.2 Results of combustion of different fuels
Property
Candle wax
Butane
mass of ‘burner’ before heating (g)
 23.77
 32.72
 43.94
 22.54
 32.50
 43.71
200
200
200
initial temperature of water (°C)
 20.0
 20.0
 20.0
highest temperature of water (°C)
 35.0
 30.0
 29.0
G
Ethanol
PA
Most of our energy is supplied by
fossil fuels.
PR
Apparatus for testing the
energy output of candle wax
mass of ‘burner’ after heating (g)
mass of fuel used (g)
TE
D
mass of water (g)
temperature rise (°C)
U
N
C
O
R
R
EC
molar mass (g mol–1)
56
Unit 3
Chapter review
FS
■■
O
The unit of energy is the joule (J).
The study of energy changes that accompany
chemical reactions is called thermochemistry or
chemical thermodynamics.
■■ The total energy stored in a substance is called the
enthalpy, or heat content, of the substance and is
represented by the symbol H.
■■ The change in enthalpy as a reaction proceeds is
known as the heat of reaction and can be determined
according to the relationship:
■■
–– Volume is measured in cubic metres (m3), litres (L)
or millilitres (mL).
1 m3 = 103 L = 106 mL
–– Quantity is measured in moles (mol).
■■ Boyle’s law states that the volume of a fixed mass of
gas at constant temperature is inversely proportional
to the pressure exerted on it.
P1V1 = P2V2
■■ Charles’ law states that the volume of a fixed mass of
gas at constant pressure is directly proportional to its
absolute (Kelvin) temperature.
O
Summary
Exothermic reactions release energy to their surroundings and have a negative ∆H value.
■■ Endothermic reactions absorb energy from their surroundings and have a positive ∆H value.
■■ Energy diagrams or profiles may be used to visually
represent changes in enthalpy.
■■ Thermochemical equations are chemical equations
that, in addition to balancing charge and mass,
include the enthalpy change and may be used in
stoichiometric calculations to determine the energy
changes associated with chemical reactions.
■■ The kinetic molecular theory of gases helps explain
gas properties and begins with five assumptions
about gas particles. These particles:
–– are moving constantly and at random
–– experience an increase in kinetic energy and move
more quickly when temperature is increased
–– have insignificant attractive or repulsive forces
between them
–– are very far apart and their volume is small compared to the volume they occupy
–– collide with one another and the walls of their container, exerting pressure.
■■ When considering gas behaviour, we are concerned
with the pressure a gas exerts, the volume it occupies,
its temperature and the number of gas molecules. We
express these in SI units.
–– Pressure is measured in pascal (Pa).
PR
∆H = Hproducts – Hreactants
V1 V2
=
T1 T2
■■
Gas volumes are measured using a standard set of
fixed conditions, standard laboratory conditions
(SLC), where temperature is 25 °C (298 K) and pressure is 100.0 kPa.
■■ The molar gas volume is the volume that 1 mole of
gas occupies. At SLC, this equals 24.8 L. To calculate
amounts of gases at SLC, we use the equation:
V
nSLC =
where V is measured in litres.
24.8
■■
U
N
C
O
R
R
EC
TE
D
PA
G
E
■■
760 mmHg = 1 atm = 1.013 × 105 Pa
= 101.3 kPa
–– Temperature is converted from degrees Celsius
(°C) to the absolute or Kelvin scale, where absolute
zero is –273 °C.
K = °C + 273
The universal gas equation combines several of
the gas laws and contains the universal gas constant (R).
PV = nRT where the constant R is equal to
8.31 J K–1 mol–1 and V is measured in litres,
T in kelvin and P in kPa.
The universal gas equation enables stoichiometric
calculations that link mases and volumes together
(mass–volume calculations).
■■ Volume–volume stoichiometry can be performed
using volumes rather than moles, since, for gases at
the same pressure and temperature, mole ratios are
equal to volume ratios.
■■
V (unknown) coefficient of unknown
=
V (known)
coefficient of known
■■ ΔH values in thermochemical equations can be
evaluated from experimental data and from the ΔH
values of related equations.
■■ The environmental effect of a fuel can be measured
in terms of its greenhouse gas emissions per megajoule of energy produced.
■■ The specific heat capacity (symbol c) of a substance is
defined as the amount of energy needed to raise the
temperature of 1 g of the substance by 1 °C (or 1 K).
CHAPTER 2 Energy calculations
57
The heat energy in a fuel can be estimated by using a
known amount of the fuel to heat a known amount of
water. The formula Q = mcΔT is used to calculate the
heat energy added to the water.
Multiple choice questions
1. Which of the energy profiles below represents the
most exothermic reaction?
with 2 moles of chlorine gas:
A 184 kJ of energy is released
B 184 kJ of energy is absorbed
C 368 kJ of energy is released
D 368 kJ of energy is absorbed.
3. Which of the following processes is
endothermic?
A Burning carbon
B Evaporating kerosene
C Condensing steam
D A reaction with a negative heat of reaction
4. ‘Dry ice’ is solid carbon dioxide. It is stable at
very low temperatures but sublimes at room
temperature according to the reaction:
CO2(s) + heat
O
Energy
FS
A
When 2 moles of hydrogen gas react completely
CO2(g)
O
■■
PR
Handling dry ice with bare hands can cause severe
TE
D
PA
Energy
G
E
B
skin damage because solid carbon dioxide:
A is a strong oxidising agent
B releases considerable heat to the skin while
subliming
C absorbs considerable heat from the skin while
subliming
D forms a strong acid when dissolved in the
moisture of the skin.
5. The diagram below shows a container filled with
gas and sealed by a movable piston. There is
sufficient gas to support the piston at point X when
the temperature is 20 °C.
movable
Assume that the piston is
piston
locked at X. If the gas is
heated, the pressure which
it exerts will:
A increase
X
B decrease
gas at
Y
20 °C
C remain the same
D change in an
container
unpredictable manner.
6. The apparatus in question 5 remains at a
constant temperature of 20 °C, but the piston
is pushed down to point Y and locked, so
that the volume of the container is halved.
The average number of molecules striking a
unit area of the wall of the container per unit
time will:
A double
B halve
C remain the same
D change in an unpredictable manner.
7. The pressure of the gas inside a scuba diver’s lungs
changes from 100 kPa to 150 kPa. If the diver’s lungs
initially held 6 L of gas, their volume at this depth
would be:
A 3 L
C 6 L
B 4 L
D 9 L.
O
R
R
Energy
EC
C
Energy
U
N
C
D
2 . Consider the following equation for the formation
of hydrogen chloride gas.
H2(g) + Cl2(g)
58
Unit 3
2HCl(g) ∆H = −184 kJ mol−1
  8. A rigid container holds a fixed volume of gas
C Heat is released to the environment by the
reaction.
D The volume occupied by the gaseous products
G
E
PR
O
O
FS
is equal to that of the reactants, if all are
measured at the same temperature and
pressure.
15. Consider the following reaction.
2H2O(g) ∆H = −484 kJ mol−1
2H2(g) + O2(g)
Energy is released when hydrogen burns in
oxygen because:
A the net strength of the chemical bonds within
the reactant molecules is greater than the net
strength of the chemical bonds within the
product molecules
B the net strength of the chemical bonds within
the reactant molecules is less than the net
strength of the chemical bonds within the
product molecules
C there are fewer product molecules than
reactant molecules
D the reactants are elements while the product
is a compound.
16. Consider the following reaction.
2H2(g) + O2(g)
U
N
C
O
R
R
EC
TE
D
PA
at a certain temperature and pressure. In order
to double the pressure of the gas inside the
container, one could:
A halve the amount of gas in the container
B halve the amount of gas, but double the
absolute temperature
C double the amount of gas and double the
absolute temperature
D double the absolute temperature.
  9. 6.0 L of air at 400 K and 100 kPa is cooled to
300 K at constant pressure. The new volume
will be:
a 2.5 L
c 6.0 L
b 4.5 L
d 8.0 L.
10. Two litres of helium gas was measured at
a temperature of 200 K and a pressure of
1 atmosphere. If the kinetic energy (temperature)
of the molecules is doubled while the pressure
remains constant, what volume will the gas
occupy?
A 0.5 L
C 2 L
B 1 L
D 4 L
11. 10 L of hydrogen gas is collected at 110 kPa and
20 °C. It is then compressed into a 2 L container at
a pressure of 1100 kPa. The new temperature will
be:
a 40 °C
c 273 °C
b 200 °C
d 313 °C.
12. 200 mL of gas at 27 °C and 700 mmHg pressure
became 210 mL at 1.05 atm pressure. The new
temperature is:
A 360 °C
C 187 °C
B 87 °C
D 260 °C.
13. Tired of working in laboratories that are either
freezing or too hot, Jenny has proposed a third set
of standard conditions. She has defined these as
a temperature of 15 °C and a pressure of 100 kPa.
Under these conditions, the molar volume of a gas
would be:
a less than 22.7 L
b 22.7 L
c between 22.7 L and 24.8 L
d greater than 24.8 L.
14. Methane, CH4(g), burns in air to form carbon
dioxide and water according to the following
equation.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
∆H = −889 kJ mol−1
Which of the following statements about this
reaction is false?
A The reaction is exothermic.
B The enthalpy (heat content) of the products
is greater than the reactants, as shown by the
sign of ∆H.
2H2O(g) ∆H = −484 kJ mol−1
Compared with the reaction given above, the heat
evolved in the reaction
2H2(g) + O2(g)
2H2O(l)
will necessarily be:
A greater, because extra heat is evolved in
forming liquid water from gaseous water
B less, because extra heat is absorbed in
forming liquid water from gaseous water
C the same, since the heat evolved is the
same provided the chemical formulas
of the reactants and products remain
unchanged
D the same, provided the same amounts of
reactants are used.
Questions 17 and 18 refer to the following
thermochemical equation for the combustion of
ethyne, commonly known as acetylene.
2C2H2(g) + 5O2(g)
4CO2(g) + 2H2O(l)
ΔH = –2600 kJ mol–1
17. The combustion of 0.26 g of ethyne will:
a absorb 13 J of energy
b evolve 13 J of energy
c absorb 13 kJ of energy
d evolve 13 kJ of energy.
18. The carbon dioxide contribution from the use of
ethyne is:
a 17 kg MJ–1
b 34 kg MJ–1
c 68 kg MJ–1
d 134 kg MJ–1.
CHAPTER 2 Energy calculations
59
(d)Calculate ΔH for the following reaction.
19. The amount of energy required to raise 150 g of
water by 40 °C is:
a 25 J
c 25 kJ
b 150 J
d 150 kJ.
20. If burning 4.5 g of a fuel raises the temperature of
75 g of water by 30 °C, the heat of combustion of
this fuel is:
a 0.0075 kJ g–1
c 2.1 kJ g–1
–1
b 0.057 kJ g
d 2.4 kJ g–1.
2CO(g) + O2(g)
2CO2(g).
7. Data tables give the heat output from the complete
O
R
R
EC
TE
D
FS
O
O
PR
E
PA
Energy changes in chemical reactions
1. What is meant by the ‘enthalpy change’ of a
chemical reaction?
2 . State whether each of the following reactions is
exothermic or endothermic.
2HCl(g)
(a)H2(g) + Cl2(g)
∆H = –185 kJ mol−1
(b)H2(g) + CO2(g)
CO(g) + H2O(g)
∆H = +41 kJ mol−1
(c)SnO2(s) + 2C(s)
Sn(s) + 2CO(g)
∆H = +360 kJ mol−1
(d)N2(g) + 3H2(g)
2NH3(g)
∆H = –92.2 kJ mol−1
3. Draw an energy profile diagram for each of
the reactions in question 2. In each diagram,
label the reactants and the products as given
by the equation, and show the ΔH value and its
magnitude.
4. Compare endothermic and exothermic reactions
with respect to:
(a) the sign of ∆H
(b) heat exchange with the surroundings
(c) energy levels of reactants and products.
5. Energy is lost during an exothermic reaction.
Explain why this observation is consistent with the
Law of Conservation of Energy.
G
Review questions
combustion of ethane and ethene as 51.9 kJ g–1
and 50.3 kJ g–1 respectively. Write thermochemical
equations for the complete combustion of these
fuels, showing ΔH values in units of kJ mol–1.
8. Dodecane has the formula C12H26. In an
experiment, 3.10 g of dodecane was burned in a
plentiful supply of oxygen. 137.0 kJ of heat was
released. Write the thermochemical equation for
the complete combustion of dodecane.
9. While researching on the internet, a student found
that candle wax is a mixture of hydrocarbons, but,
as an approximation, it may be assumed to be
pentacosane.
(a) To what homologous series does pentacosane
belong?
(b) Write the formula for pentacosane given that
its molecules contain 25 carbon atoms.
(c) Write the equation for the combustion of
pentacosane in a plentiful supply of oxygen.
In an experiment, the student measured the heat
given off by a burning candle. She discovered
that 153 kJ of energy was evolved and that the
mass of the candle decreased from 75.384 to
71.735 g.
(d) On the assumption that candle wax may be
approximated as pentacosane, determine
the heat of combustion in kJ mol–1 and then
write the thermochemical equation for its
combustion.
On further investigation, she found a source
claiming tetracosane, C24H50, was a better
approximation for candle wax.
(e) How would this change affect the value of ΔH?
U
N
C
Writing thermochemical equations
6. Ethanol burns in oxygen to produce water and
either carbon dioxide or carbon monoxide.
The particular oxide produced depends on
whether the oxygen supply is plentiful or limited.
The molar heats of combustion for these two
reactions are –1364 kJ mol–1 and –1192 kJ mol–1
respectively.
(a) Write the thermochemical equation for the
combustion of ethanol to produce carbon
dioxide.
(b) Write the thermochemical equation for the
combustion of ethanol to produce carbon
monoxide.
(c) Use the equations from (a) and (b) to explain
how the amount of oxygen consumed
influences the oxide produced.
60
Unit 3
Measuring gases
10. Use the kinetic molecular theory to explain:
(a) why gases exert pressure
(b) why, for a fixed amount of gas at
constant volume, pressure increases with
temperature.
11. (a) Convert 1.34 atm to mmHg and kPa.
(b) Convert 365 mmHg to Pa and atm.
(c) Convert 102 576 Pa to atm and mmHg.
12. Convert the following Celsius temperatures to
Kelvin temperatures.
(a) 100 °C
(c) –200 °C
(b) –20 °C
(d) 345 °C
13. Convert the following Kelvin temperatures to
Celsius.
(a) 300 K
(d) 392 K
(b) 427 K
(e) 73 K
(c) 173 K
14. Convert the following volumes to the second unit
given.
(a) 125 L to m3
(b) 125 L to mL
(c) 300 mL to L
O
O
PR
E
G
PA
Standard conditions
15. Calculate the number of moles of the following
gases at SLC.
(a) 1.5 L oxygen, O2
(b) 2.56 L of hydrogen, H2
(c) 250 mL of nitrogen, N2
16. Calculate the volume of the following gases at
SLC.
(a) 1.53 mol hydrogen, H2
(b) 13.6 g of methane, CH4
(c)2.5 × 1030 molecules of nitrogen, N2
17. Calculate the mass of the following gas samples.
All volumes are measured at SLC.
(a) 150 mL of oxygen, O2
(b) 4.5 L of carbon dioxide, CO2
18. (a)How many moles of nitrogen are contained in
26.0 L of nitrogen gas at SLC?
(b) How many individual nitrogen molecules are
present in 26.0 L of the gas?
FS
(d) 2.6 mL to L
(e) 2 m3 to L
(f ) 3 × 103 mL to L
N
C
O
R
R
EC
TE
D
Gas laws
19. What is absolute zero? Can one actually reach this
temperature? Explain.
20. The pressure on 13.0 L of neon gas is increased
from 90.0 kPa to 360 kPa. If the temperature
remains constant, find the new volume of
the gas.
21. 1.00 L of hydrogen gas in a cylinder under
5.0 × 104 kPa fills a balloon at 150 kPa at the
same temperature. Determine the volume of the
balloon.
22. The pressure on a gas remains constant. Its
volume is 700 mL. The temperature is 27 °C.
Calculate the temperature needed to change the
volume to:
(a) 14.0 mL
(b) 420.0 mL.
U
The universal gas equation
23. Calculate the number of moles of gas present in
each of the following gas samples.
(a) 32.3 L of argon at 102.0 kPa and 15 °C
(b) 24.3 L of nitrogen at 13.2 atm and 35 °C
(c) 12.3 L of helium at 755 mmHg and –10 °C
(d) 24.8 L of argon at 100.0 kPa and 25 °C
24. Calculate the volumes occupied by the following
gas samples.
(a) 0.560 mol of carbon dioxide gas at 101.5 kPa
and 17 °C
(b) 15.5 mol of ethyne gas at 10.35 atm and 20 °C
(c) 0.800 g of oxygen gas at 98.9 Pa and 11 °C
(d) 5.279 g of nitrogen gas at 100 kPa and –50 °C
25. An aerosol can of deodorant has a volume
of 120 mL. The contents exert a pressure of
9.0 × 105 Pa at 27 °C.
(a) Calculate the number of moles of gas present
in the can.
(b) How many particles are present in the can of
deodorant?
(c) If the contents of the can are transferred
to a 200 mL container, what would be
the temperature if the pressure drops to
6.0 × 105 Pa?
26. Calculate the temperature, in °C, of 0.25 mol of a
gas that occupies a volume of 4.155 × 10–3 m3 at a
pressure of 150 kPa.
27. A certain mass of unknown gas (molar mass =
30.0 g mol–1) occupies 16.64 L at a pressure of
125 mmHg and a temperature of –73 °C. Calculate
the mass of gas present in the container.
28. Calculate the pressure needed to contain 3.0 mol
of ammonia in a container of volume 20 L at a
temperature of 27 °C.
29. An empty 200 mL flask has a mass of 84.845 g. It is
filled with a gas at 17.0 °C and 770 mmHg pressure
and then weighs 85.084 g. Calculate the molar
mass of the gas. What is the gas?
30. A gas is at SLC and has a volume of 11.2 L. If
its temperature is changed to 127 °C, and its
pressure changes to 253.3 kPa, calculate its new
volume.
31. Each time Nicole breathes, she inhales about
400 mL of air. Oxygen makes up about 20% by
volume of air. How many oxygen molecules does
she inhale in one breath at 25 °C and 1.0 × 104 Pa
pressure?
Mass–volume calculations
32. Magnesium burns according to the equation:
2Mg(s) + O2(g)
2MgO(s)
What mass of magnesium combines with 5.80 L of
oxygen measured at SLC?
33. Butan-1-ol (density = 0.81 g mL–1) burns according
to the following equation.
CH3CH2CH2CH2OH(l) + 6O2(g)
4CO2(g) + 5H2O(l)
When 10.0 mL of butan-1-ol is burned, calculate:
(a) the mass of water produced
(b) the volume of carbon dioxide produced at SLC
(c) the volume of carbon dioxide produced at
200 °C and 1.2 atm pressure.
The butan-1-ol is used to produce 100 mL of
carbon dioxide at SLC.
(d) Calculate the volume of butan-1-ol needed.
CHAPTER 2 Energy calculations
61
400 °C

→
CaO(s) + CO2(g) steam
CaCO3(s)
(a) Calculate the maximum volume of carbon
dioxide (at SLC) that can be removed per
1.00 tonne of calcium oxide.
(b) Calculate the maximum mass of calcium
carbonate that would be formed per tonne of
calcium oxide.
2H2(g) + O2(g)
O
PR
E
G
2H2O(g)
What volume of water vapour is produced when
4.0 L of hydrogen reacts with 10 L of oxygen? What
volume of oxygen remains unreacted? (Assume
that all volumes are measured at the same
temperature and pressure.)
38. In the Haber process, nitrogen and hydrogen
are combined under specific conditions of
temperature and pressure to industrially
manufacture ammonia.
(a) Write the balanced equation.
(b) What volume of nitrogen combines with 45 m3
of hydrogen?
(c) What volume of ammonia would be produced
if the reaction went to completion?
U
N
C
O
R
R
Calculations using thermochemical equations
39. Iron can react with carbon dioxide according to
the following equation.
2Fe(s) + 3CO2(g)
62
Unit 3
SO2(g) ∆H = –296 kJ mol−1
(a) how much heat is evolved when 2.00 moles of
sulfur are burned in excess oxygen
(b) how much heat is evolved when 200 g of sulfur
is burned in excess oxygen
(c) how much heat is evolved when 450 mL of
oxygen at SLC reacts with excess sulfur
(d) how much heat is evolved when 200 g of sulfur
reacts with 2 litres of oxygen at 80 °C and
101.3 kPa pressure?
41. Given the following two equations:
PA
EC
TE
D
Volume–volume stoichiometry
35. Methane gas, CH4, is a major constituent of the
natural gas in Bass Strait. It readily burns in
air to produce carbon dioxide and water vapour. If
all gas volumes are measured at SLC calculate:
(a) the volume of oxygen gas that is required
for the complete combustion of 12.5 L of
methane gas
(b) the volume of air required for reaction, given
that air contains 20% oxygen.
36. When propane, C3H8, is burned in a limited supply
of oxygen, carbon monoxide is among its products.
If 4.25 L of propane gas, measured at room
temperature and pressure, is burned in a limited
air supply, what volume of carbon monoxide gas
would be produced at the same conditions?
37. Hydrogen reacts with oxygen in the reaction:
S(s) + O2(g)
FS
(a) Explain why this equation may be described
as a thermochemical equation.
(b) Calculate the heat change, in kilojoules, when
2 g of iron reacts completely with excess
carbon dioxide.
(c) Calculate the heat change, in kilojoules,
when 200 mL of carbon dioxide is used at
SLC.
(d) What mass of iron would be needed to absorb
1000 kJ of heat energy?
(e) What volume of carbon dioxide would be
needed at SLC to absorb 1000 kJ of energy?
40. For the reaction:
developed for removing carbon dioxide from flue
gases. Some of these are also being investigated
for the removal of carbon dioxide directly from
the air. One such process uses calcium oxide and
steam (at 400 °C) to remove carbon dioxide. The
overall equation for this reaction is:
O
34. A number of technologies either exist or are being
Fe2O3(s) + 3CO(g)
∆H = +26.3 kJ mol−1
1
N (g)
2 2
1
+ 2 O2(g)
2NO(g) + O2(g)
NO(g) ΔH = 90 kJ mol–1
2NO2(g) ΔH = –114 kJ mol–1
calculate the enthalpy change for the following
reaction:
2NO2(g)
N2(g) + 2O2(g)
42. Butane, C4H10, is a gaseous fuel. For the
combustion of butane, ∆H = −2874 kJ mol−1:
(a) write a balanced thermochemical equation
(b) calculate the quantity of heat that would be
evolved from the combustion of 12 L of butane
measured at SLC.
43. Hydrazine, N2H4, is a liquid fuel that has been used
for many years in the engines of space probes.
It was famously used in the terminal-descent
engines that successfully landed the Curiosity
rover on the surface of Mars in 2012. When passed
over a suitable catalyst, it decomposes quickly in
a multi-step exothermic chemical reaction. The
overall equation for this process is:
N2H4(l)
N2(g) + 2H2(g) ΔH = –50.3 kJ mol–1
(a) Calculate the energy released per kilogram of
hydrazine in the equation above.
(b) Given that the average temperature and
pressure of the Martian atmosphere are
–60 °C and 600 Pa respectively, calculate the
total volume of gas added to the Martian
atmosphere if 50 kg of hydrazine is used.
propane
–2217
Density
(g mL –1)
0.51 (as LPG)
butane
–2874
0.51 (as LPG)
octane
–5464
0.70 (as petrol)
FS
G
Molar enthalpy of
combustion (kJ mol –1)
PA
Fuel
E
(a) Given that the density of octane is 0.70 g mL–1,
calculate the volume of carbon dioxide
produced (at SLC) per litre of octane.
(b) Calculate the mass of carbon dioxide evolved
in the production of 1.00 MJ of energy through
the reaction above.
(c) Express your answer to (b) as litres of carbon
dioxide (at SLC) per MJ.
45. Petrol and LPG are two fuels commonly used
in Australia. It is claimed that LPG is ‘better
for the environment’ as it releases less carbon
dioxide. It is also attractive to motorists because,
even though more litres are used, it is cheaper
than petrol. As an approximation, petrol may be
assumed to be octane, whereas LPG is a mixture
of propane and butane. Some relevant data is
shown in the table below.
45.6 g of lead from 19.2 °C to 32.5 °C. Calculate the
specific heat capacity of lead.
48. A student investigated the heat output from some
solid fuel blocks designed for use with model
steam engines. In one experiment, it was found
that the temperature of 80.0 mL of water increased
from 23.2 °C to 68.5 °C, and 0.300 g of fuel was
burned in the process. Calculate the heat content
of this fuel in kJ g–1.
49. It is useful to know how much energy can be
obtained from different fuels in order to determine
which would be the best fuel for a particular
purpose. The apparatus in the figure below can
be constructed in the laboratory to measure the
heat given out when a fuel such as ethanol is
burned.
The heat produced when the fuel burns is
absorbed by the water in the metal can. The
temperature can be measured so, given that the
specific heat of water is 4.18 J and the density of
water is 1.00 g mL–1, the heat of combustion may
be determined.
O
16CO2(g) + 18H2O(l)
ΔH = –10 928 kJ mol–1
O
2C8H18(l) + 25O2(g)
47. It takes 78.2 J of energy to raise the temperature of
PR
44. Octane burns according to the following equation.
small metal
can
liquid
burner
U
N
C
O
R
R
EC
TE
D
(a) Calculate the mass of LPG (assuming it to be
propane) required to produce 1.00 MJ of heat
energy.
(b) Calculate the mass of carbon dioxide produced
from (a) and express your answer as g MJ–1.
(c) Calculate the mass of petrol (assuming it to be
octane) required to produce 1.00 MJ of heat
energy.
(d) Calculate the mass of carbon dioxide produced
from (c) and express your answer as g MJ–1.
(e) Hence state the net reduction (in g MJ–1) of
carbon dioxide emission when LPG is used in
preference to petrol.
(f ) Repeat (a), (b) and (e) for LPG if it is assumed
to be butane.
(g) Calculate the volume of LPG (assuming it to
be propane) required to produce the same
energy as 1.00 litre of petrol (assuming it to be
octane).
(h) Is it true that LPG is better than petrol?
Use your answers to (a)–(g) to explain your
response.
measured
volume of
water
Estimating heat energy in fuels
46. The specific heat capacity of aluminium is
0.900 J °C−1 g−1. Calculate the energy needed to
raise the temperature of a 7.20 × 102 g block of
aluminium by 10.0 °C.
methylated
spirits
Apparatus for measuring the
heat of combustion of a fuel
CHAPTER 2 Energy calculations
63
(d) Outline the sources of error in the experiment
and then suggest how the design of the
experiment could be improved so that more
accurate heats of combustion for different
fuels may be determined.
50. A student wished to estimate the ΔH value for the
reaction:
The results of one experiment are shown in the
following table.
Volume of water in metal can =
200 mL
Thus, mass of water in can =
200 g
Rise in temperature of water =
11.0 °C
Mass of ethanol burned =
0.500 g
C5H12(l) + 8O2(g)
(a) From these results, calculate the heat
produced when 1 gram of ethanol is
burned.
(b) Calculate the heat of combustion for ethanol
using the above results.
(c) An accurate value for the heat produced when
1 mole of ethanol burns is 1364 kJ mol–1.
Calculate the percentage accuracy of this
experiment.
using an identical experimental set-up to that in
question 49. Using pentane in the liquid burner, it
was found that the combustion of 0.735 g caused a
temperature increase of 35.0 °C.
(a) Estimate the ΔH value for the equation above
using the data obtained.
(b) Given that the molar enthalpy of combustion
of pentane is –3509 kJ mol–1, comment on the
value obtained in (a).
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
64
Unit 3
5CO2(g) + 6H2O(l)
Exam practice questions
Sit VCAA exam
In a chemistry examination, you will be required to answer a number of multiple choice
and extended response questions.
Extended response questions
1. Nitroglycerine, C3N3H5O9, is a dangerous explosive that releases all gaseous products according to the
equation:
6N2(g) + 12CO2(g) + 10H2O(g) + O2(g)
FS
4C3N3H5O9(l)
3 marks
2 marks
2CO2(g) + 4H2O(l) ΔH = –1450 kJ mol–1
2CH3CH2CH2OH(l) + 9O2(g)
PA
2CH3OH(l) + 3O2(g)
1 mark
G
E
PR
O
O
Calculate the total volume of gaseous products that results when 1.00 kg of liquid nitroglycerine
explodes at a temperature of 250 °C and a pressure of 300 kPa.
2.Kerosene is a hydrocarbon fuel that may be used in lamps, jet engines and camp stoves. It has a
heat of combustion of 44 100 kJ kg–1.
(a) Explain why the heat evolved from the combustion of kerosene is measured in kJ kg–1 rather
than kJ mol–1.
(b) A cup of billy tea contains 250 g of water. How many cups of tea can be made if 12.5 mL of
kerosene is used to heat the water? (Assume that the temperature of the water increases
from 20.0 °C to 100.0 °C, the specific heat capacity of water is 4.18 J g–1 °C–1 and the heat of
combustion of kerosene is 37 000 kJ L–1.)
3.Consider the following two equations.
6CO2(g) + 8H2O(l) ΔH = –4032 kJ mol–1
2 marks
2 marks
(a) Calculate the mass of methanol required to produce 1.00 MJ of heat energy.
D
(b) Calculate the mass of 1-propanol required to produce 1.00 MJ of heat energy.
(c) Assuming that methanol and 1-propanol are made from non-renewable resources, calculate
3 marks
3 marks
U
N
C
O
R
R
EC
TE
the net mass reduction (in g MJ–1) of carbon dioxide when methanol is used as a fuel in
preference to 1-propanol.
(d) Calculate the net volume reduction (in L MJ–1) from (c), assuming the carbon dioxide is at
101.3 kPa and 15 °C.
CHAPTER 2 Energy calculations
65